E7.3 Magnitude of a vector

Learning objectives

E7.3.1Calculate the magnitude of a vector x y J L KK N P OO as xy22+• Calculate the magnitude of a vector x y J L KK N P OO as xy22+. The magnitudes of vectors will be denoted by modulus signs, e.g. • a is the magnitude of a • AB is the magnitude of AB.

The magnitude of a vector is its non-negative length. For a two-dimensional vector, the horizontal and vertical components form the perpendicular legs of a right triangle, so Pythagoras gives the length.

\left|\begin{pmatrix}x\y\end{pmatrix}\right|=\sqrt{x^2+y^2}

The magnitude of $\mathbf a$ is written $|\mathbf a|$ , and the magnitude of $\overrightarrow{AB}$ is written $|\overrightarrow{AB}|$ or $AB$ . Direction changes the vector but not its magnitude, so $|\overrightarrow{AB}|=|\overrightarrow{BA}|$ .

For $\mathbf v=\begin{pmatrix}-4\\5\end{pmatrix}$ , $|\mathbf v|=\sqrt{(-4)^2+5^2}=\sqrt{41}\approx6.40$ . Keep $\sqrt{41}$ when an exact value is required; otherwise round only the final value to the requested accuracy.

The same relationship can find an unknown. If $\overrightarrow{MT}=\begin{pmatrix}2k\\-k\end{pmatrix}$ and $|\overrightarrow{MT}|=\sqrt{180}$ , then $\sqrt{5k^2}=\sqrt{180}$ , so $|k|=6$ ; if $k$ is stated positive, $k=6$ .

Square each signed component before adding: $(-4)^2=16$ , not $-16$ . Magnitude is a scalar, never negative, and it has no direction. Do not add the components first or omit the square root.

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—E7.4 Vector geometryRead